aa pearson maths 9 sb-03.fm page 167 monday, october 24...

Half-time 3

3 Algebra 167

1 Simplify each of the following, expressing your answer with positive indices.

(a) 3x3y × 4x4y (b) (c)

2 Rearrange the formula z = 2xy − 3y to make x the subject and then find the value of x when z = 7 and y = 5.

3 Simplify each of the following, expressing your answer with positive indices.

(a) (3x4y5)2 (b) (c)

4 Express each quantity described below in scientific notation.

(a) The Moon is 384 000 km away.

(b) 500 mL of water is lost through the skin daily.

5 To how many significant figures have each of the following been written?

(a) 36 050 000 (b) 0.000 000 852 (c) 0.0830

6 Simplify the following, leaving your answer in index form.

(a) (b) (c)

7 Express each quantity described below as a number.

(a) The number of bacteria that will fit on a pin head is approximately 1.1 × 1014.

(b) The nucleus of a leaf cell is about 4.4 × 10−6 m in diameter.

8 Simplify each of the following, expressing your answer with positive indices:

(a) (b)

(c) (d)

9 Express each of the following with positive indices.

(a) x−5 (b) 3p−4 (c) 5m−3n (d) 6p−4q−1r2

10 Write the following numbers correct to three significant figures:

(a) 0.2999 (b) 80 (c) 62 078

3.18x3y5

2xy2-------------- 3m4n2 2m5× n3

8m3n4---------------------------------------

3.4

3.2a4b3

6ab4-----------⎝ ⎠

⎛ ⎞ 2 53x4y6

3x3y2----------------

x6yx7y5----------×

3.3

3.3

3.123( )4 56( )× 2

25( ) 52( )4×-------------------------------- 32 54×( )2

25 27×------------------------ 83 92×

64-----------------

3.3

3.2p2q3( )-1 p5q-2×

p3q2---------------------------------------- c-3d-4( )2 c2d6×

c5d3---------------------------------------

4 a2b( )3 3a-5b-2×6ab

-------------------------------------------- 6 mn3( )2 4m-4n-1×3m2n

------------------------------------------------

3.2

3.3

AA Pearson Maths 9 SB-03.fm 67 Monday, October 24, 2011 10:19 AM

168 PEARSON mathematics 9

Expanding the brackets

Brackets are used to change the order of operations, allowing addition or subtraction to be done before multiplication or division. We can write the expression without brackets

if we multiply every term inside the brackets by the number outside the brackets. This is called expanding the brackets.

2(3 + 4) = 2 × 7 and 2(3 + 4) = 2 × 3 + 2 × 4 = 14 = 6 + 8

= 14

If we replace numbers with variables, we can rewrite the expression without the brackets the same way.

This is known as the distributive law. Every term inside the brackets is multiplied by the term outside the brackets.

Distributive law

a(b + c) = ab + ac

Worked Example 15Expand:

(a) 5(p + 4) (b) 2m(4m − 3n)

Thinking(a) Multiply all terms inside the brackets

by the term in front of the brackets.

(Here 5(p + 4).)

(a) 5(p + 4) = 5 × p + 5 × 4= 5p + 20

(b) Multiply all terms inside the brackets by the term in front of the brackets.

(Here 2m(4m − 3n).)

(b) 2m(4m – 3n) = 2m × 4m + 2m × -3n= 8m2 – 6mn

Area = 14 units2 2 units 6 units2 8 units2 2 units

3 units 4 units3 units 4 units

Area = a(b + c)

a(b + c)

a Area = ab Area = aca

=

b cb units c units

ab ac+

15

3.5AA Pearson Maths 9 SB-03.fm Page 168 Monday, October 24, 2011 10:19 AM

3 Algebra

3.5

169

Binomial productsA binomial expansion is the result of multiplying two terms in one set of brackets by two terms in another set of brackets.

(a + b)(c + d) = ac + ad + bc + bd.

Consider a room 6 metres long and 4 metres wide. If it is extended by 3 metres in length and 2 metres in width, find the new area, as in the diagram below.

The total area can be found in two ways:

(6 + 3)(4 + 2) = 9 × 6 or 6 × 4 + 6 × 2 + 3 × 4 + 3 × 2 = 24 + 12 + 12 + 6 = 54 m2 = 54 m2

Worked Example 16Expand and simplify:

(a) 4a(a − 3) − 2(a + 5) (b) 3(2 − b) − (4 − b)

Thinking(a) 1 Expand both brackets by multiplying

by the term in the front of the brackets. (Here we multiply the first bracket by 4a and the second bracket by -2.)

(a) 4a(a − 3) − 2(a + 5)= 4a × a + 4a × (-3) + (-2) × a + (-2) × 5= 4a2 − 12a − 2a − 10

2 Simplify by collecting like terms. = 4a2 − 14a − 10

(b) 1 Expand both brackets by multiplying by the term in the front of the brackets. (Here we multiply the first bracket by 3 and the second bracket by -1. Remember that the product of two negatives is positive: -1 × -b = b.)

(b) 3(2 − b) − (4 − b)= 3(2 − b) − 1(4 − b)= 3 × 2 + 3 × (-b) + (-1) × 4 + (-1) × (-b)= 6 − 3b − 4 + b

2 Simplify by collecting like terms. = 2 − 2b

16

6 m

6 m

9 m

3 m

4 m

2 m

Area = 24 m2

Area = 12 m2

Area = 12 m2

Area = 6 m2

AA Pearson Maths 9 SB-03.fm Page 169 Monday, October 24, 2011 10:19 AM

3.5


If a room a metres long and b metres wide has been extended by 2 metres in length and 5 metres in width, we can find the new area.

Area = (a + 2)(b + 5)= a × b + a × 5 + 2 × b + 2 × 5= ab + 5a + 2b + 10 m2

If a room a metres long and c metres wide has been extended by b metres in length and d metres in width, we can find the new area using the following diagram:

Area = (a + b)(c + d)= a × c + a × d + b × c + b × d= ac + ad + bc + bd m2

Binomial expansion

(a + b)(c + d) = ac + ad + bc + bd

Every term in the second bracket is multiplied by every term in the first bracket.

Worked Example 17Expand each of the following expressions.

(a) (c + 4)(d + 2) (b) (2m − 4)(3n − 5) (c) (2p − 3r)(3t + 4)

Thinking(a) 1 Expand the following by multiplying

every term in the second bracket by every term in the first bracket.

(Here (c + 4)(d + 2).)

(a) (c + 4)(d + 2)= c × (d + 2) + 4 × (d + 2)

= c × d + c × 2 + 4 × d + 4 × 2

2 Write without multiplication signs and perform any possible calculations.

= cd + 2c + 4d + 8

(b) 1 Expand the following by multiplying every term in the second bracket by every term in the first bracket.

(Here (2m − 4)(3n − 5).)

(b) (2m − 4)(3n − 5)= 2m × (3n − 5) + (-4) × (3n − 5)= 2m × 3n + 2m × (-5) + (-4) × 3n + (-4)

× (-5)


= 6mn − 10m − 12n + 20

(c) 1 Expand the following by multiplying every term in the second bracket by every term in the first bracket.

(Here (2p − 3r)(3t + 4).)

(c) (2p − 3r)(3t + 4)= 2p × (3t + 4) + (-3r) × (3t + 4)= 2p × 3t + 2p × 4 + (-3r) × 3t + (-3r)

× 4

a m 2 m

(a + 2) m

(b + 5) m

b m

5 m

Area = ab m2 Area = 2b m2

Area = 5a m2 Area = 10 m2

a m b m

c m

d m

(a + b) m

(c + d) m

Area = ac m2 Area = bc m2

Area = ad m2 Area = bd m2

17


3 Algebra

3.5

171

Binomial expansion can be completed using FOIL.

First: Multiply the first terms in each bracket.Outside: Multiply the outside terms in each bracket.Inside: Multiply the inside terms in each bracket.Last: Mutiply the last terms in each bracket.

(a + b)(c + d)

Often after expansion, like terms can be collected and the expression simplified.

We are sometimes left with three terms called a quadratic trinomial. This is demonstrated in the following worked example.


= 6pt + 8p − 9rt − 12r

Worked Example 18Expand each of the following expressions. Simplify where possible.

(a) (x − 2) (x + 5) (b) (t − 3)(t − 7) (c) (4x + 3)(5x − 2)

Thinking(a) 1 Expand by multiplying every term in

the second bracket by every term in the first bracket.

(Here (x – 2)(x + 5).)

(a) (x – 2)(x + 5)

= x × x + x × 5 + (-2) × x + (-2) × 5


= x2 + 5x − 2x − 10

3 Simplify by collecting like terms. = x2 + 3x − 10

(b) 1 Expand by multiplying every term in the second bracket by every term in the first bracket.

(Here (t – 3)(t − 7).)

(b) (t − 3)(t − 7)

= t × t + t × (-7) + (-3) × t + (-3) × (-7)

2 Write without multiplication signs and perform any possible calculations. Notice that the product of two negative numbers is a positive number (-3 × -7 = +21).

= t2 − 7t − 3t + 21

3 Simplify by collecting like terms. = t2 − 10t + 21

(c) 1 Expand by multiplying every term in the second bracket by every term in the first bracket.

(Here (4x + 3)(5x – 2).)

(c) (4x + 3)(5x − 2)

= 4x × 5x + 4x × (-2) + 3 × 5x + 3 × (-2)

18


3.5


Expanding the brackets

Fluency1 Expand each of the following expressions.

(a) 2(x + 5) (b) 4(a + 9) (c) 8(b + 3)

(d) 8(x − 2) (e) 9(x − 2) (f) 3(k − 4)

(g) -7(x + 1) (h) -4(d + 8) (i) -6(x + 3)

(j) -9(c − 6) (k) -2(x − 5) (l) -5(x − 6)

(m) x(x + 3) (n) m(m + 5) (o) x(x + 9)

(p) -x(x − 8) (q) -x(x − 4) (r) -n(n − 9)

(s) 3x(x + 2) (t) 2u(u − 6) (u) 5x(x − 1)

(v) -2x(x + 7) (w) -7q(q + 1) (x) -9b(b − 6)

2 Expand and simplify each expression.

(a) 3(x − 2) + 2(x + 4) (b) 2(y + 1) + 3(y − 2)

(c) 4(2x − 3) − (6x − 1) (d) 3(4a + 5) − (7a + 2)

(e) a(2a + 3) − 2(3a − 5) (f) 3x(x + 2) − 4(x − 1)

(g) 7y(y − 3) − 5(y + 2) (h) 6k(2k + 3) − 3k(k − 3)

3 Expand each of the following.

(a) (x + 6)(y + 5) (b) (x + 4)(y + 7) (c) (c + d)(e + 3)

(d) (r + 5)(t − 8) (e) (a + 2)(b − 1) (f) (p + 3)(q − 2)

(g) (a − 3)(y − 4) (h) (m − 2)(n − 6) (i) (k − 5)(m − 4)

(j) (4p + q)(m + n) (k) (7x + y)(a + k) (l) (3p + 2r)(4s + t)

(m) (2a + 1)(b − 3) (n) (3x − 1)(x + 2) (o) (3m + 2)(n − 1)

(p) (2a − 3b)(5m − 3n) (q) (5c − 2d)(3x − 4y) (r) (7p − 2q)(2r − 3t)

4 Expand each of the following and simplify.

(a) (a + 3)(a + 2) (b) (c + 5)(c + 3) (c) (y + 7)(y + 6)

(d) (x − 3)(x + 1) (e) (y − 2)(y + 9) (f) (a − 7)(a + 12)

(g) (b − 2)(b − 1) (h) (x − 4)(x − 3) (i) (p − 8)(p − 7)

2 Write without multiplication signs and perform any possible calculations. Notice that the product of a positive and negative number is a negative number (+3 × -2 = -6).

= 20x2 − 8x + 15x − 6

3 Simplify by collecting like terms. = 20x2 + 7x − 6

NavigatorQ1, Column 1, Q2 Column 1, Q3 Column 1, Q4 Column 1, Q5 Column 1, Q6 Column 1,Q7, Q8, Q9, Q10, Q14, Q15

Q1 Column 2, Q2 Column 1, Q3 Column 2, Q4 Column 2, Q5 Column 2, Q6 Column 2, Q7, Q8, Q9, Q10, Q14, Q16

Q1 Column 3, Q2 Column 2, Q3 Column 3, Q4 Column 3, Q5 Column 3, Q6 Column 3, Q7, Q8, Q9, Q10, Q11, Q12, Q13, Q15, Q16

3.5Answers

page 609

15

Remember -(x + 2) really means -1(x + 2).

Hint

16

17

18


3

Algebra

3

.

5

173

(j)

(

3

m

+

1

)(

2

m

+

3

)

(k)

(

4

d

+

5

)(

2

d

+

3

)

(l)

(

3

k

+

5

)(

7

k

+

4

)

(m)

(

4

y

−

3

)(

2

y

+

1

)

(n)

(

5

k

+

3

)(

2

k

−

7

)

(o)

(

2

a

+

5

)(

4

a

−

3

)

(p)

(

6

a

−

5

)(

2

a

−

1

)

(q)

(

4

p

−

3

)(

3

p

−

5

)

(r)

(

3

r

−

2

)(

8

r

−

7

)

5

Expand each of the following and simplify.

(a)

(

x

+

y

)(

x

+

y

)

(b)

(

m

+

n

)(

m

+

n

)

(c)

(

a

+

b

)(

a

+

b

)

(d)

(

p

−

r

)

2

(e)

(

c

−

d

)

2

(f)

(

g

−

h

)

2

(g)

(

2

c

+

3

d

)(

2

c

+

3

d

)

(h)

(

3

m

+

2

n

)(

3

m

+

2

n

)

(i)

(

5

a

+

3

b

)(

5

a

+

3

b

)

(j)

(

5

x

−

4

y

)(

5

x

−

4

y

)

(k)

(

2

j

−

3

k

)(

2

j

−

3

k

)

(l)

(

7

c

−

5

d)2

6 Expand each of the following and simplify.

(a) (m + n)(m − n) (b) (p + q)(p − q) (c) (a − b)(a + b)

(d) (3a + 2b)(3a − 2b) (e) (4x − 3y)(4x + 3y) (f) (7m − 3n)(7m + 3n)

7 When simplified, 5m(-3 + 2n) − n(10m − 1) is:

A n − 15m B 10n − 15m − 1 C -15m − 20mn + n D 15m + n

8 (3x − 2)(3x + 2) is the same as:

A 6x2 + 4 B 9x2 − 12x − 4 C 9x2 + 6x − 4 D 9x2 − 4

Understanding9 Consider this envelope.

(a) Write down the expression (in factorised form) for the area of this envelope.

(b) Use your answer to find the area when x = 10.

(c) What value would x represent if the area of the envelope is 104 cm2? Try substituting different values for x until you find the appropriate one.

(d) Expand the expression you obtained for the area in part (a).

(e) Check that the expanded expression is correct by calculating the area when x = 10 and comparing with the answer obtained in part (b).

10 A vegetable bed, 15 m long and 3 m wide, is to be extended by x metres in both length and width.

(a) Write expressions for each of the following.

(i) The new length of the garden bed.

(ii) The new width of the garden bed.

(iii) The new area of the garden bed as the product of factors.

(iv) The new area of the garden bed as the expanded product of factors.

(b) Compare the area of the new garden bed with the area of the original garden bed. By how much has the area increased?

(c) Use your answer from part (b) to calculate how much extra area of garden bed is available if both the length and width are increased by 2 m.

(x + 7) cm

(x + 2) cm


3.5


11 The sides of a rectangle and a triangle have the lengths shown below.

(a) Write down the expression for the area of the rectangle using the given lengths.

(b) Write down the expression for the perimeter of the triangle using the given lengths.

(c) Find two values for x so that the area of the rectangle is the same as the perimeter of the triangle.

Reasoning12 There are a number of ways to check that x(x + 2) = x2 + 2x is a true statement. Choose two

of these methods and show clearly how you are able to check that x(x + 2) = x2 + 2x is true. Can you use these methods to show that a(b + c) = ab + ac? If so, show how.

13 A backyard, square in shape, is to have an in-ground rectangular swimming pool constructed in it. A path around the pool is to be laid as shown in the diagram.

(a) Write expressions for each of the following:

(i) the total area of the back yard

(ii) the width of the swimming pool

(iii) the length of the swimming pool

(iv) the area of the swimming pool.

(b) Using your answers to parts (i) and (iv), write an expression for the area of the path. State your answer in simplest expanded form.

14 Choose two variables to represent the length and width of a rectangle. Increase each dimension by a fixed amount (e.g. increase the length by 2 units and the width by 3 units). Find the area of the new rectangle.

Open-ended15 Three students gave the following incorrect answers when expanding -3(5x − 2):

(a) -15x − 2 (b) 5x + 6 (c) -8x + 5Explain the errors made by each student. What is the correct answer?

16 Three students obtained the following incorrect answers when expanding (x − 3)(x + 4):(a) x2 + 7x + 12 (b) 2x − 12 (c) x2 + x − 7Explain what errors were made by each student. What is the correct answer?

x cm

x + 3 cm

2x cm

3x cm

5x – 12 cm

pool

x

x 2 m 2 m

1 m

1 m

Problem solvingDivisibility dilemma?The number 123 456 is divisible by 2, 3 and 4. Find the smallest number that is divisible by all the digits from 2 to 9 (inclusive) if it begins with the digits 123 456.

• Test all possible combinations.

• Guess, check and improve.

• Break problem into manageable parts.

Strategy options


Technology Exploration GeoGebra

3 Algebra 175

Equipment required: 1 brain, a computer with a GeoGebra program

Versions of this exploration for other technologies are available on Pearson Reader.

Increasing profitMr Slew is employed by a company as an engineer to find a new production process. After some research, he creates a new process for the manufacture of calculators. We will use a spreadsheet, graphs and algebra to explore whether Mr Slew’s new process makes the production process cheaper, and therefore increases the company’s profits.

It is assumed that every calculator made in a day will be sold.

1 A company that makes calculators has found that the cost, $C, of producing x calculators per day is given by the rule C = -0.01(x + 300)(x − 2300), x ≤ 1500. At present, the company can produce a maximum of 1500 calculators per day. It has also found that the revenue (income for the company), $R, from the sale of x calculators is given by the rule R = -0.01x(x − 3000). What cost is incurred each day if no calculators are produced?

2 Open a GeoGebra program. You will see seven menu options (File, Edit etc.) at the top of the screen. Below these are 11 icons called tools. By clicking on the small arrow in the bottom right-hand corner of the tool icon, a drop-down list of more tools appears. The arrow turns red when you hover the cursor over it. If you hover over a tool, the tool’s name and how to use it will appear in the top right-hand corner of the screen.

Go to Options, Labeling and select ‘No new objects’.

From the menu, select ‘View’ and from the drop-down menu select ‘Spreadsheet view’. Enter Number of calculators (x)’ in cell A1, Cost C($) in B1 and Revenue R($) in C1. (Double clicking in the cell will allow you to edit your input.)

Enter values from 0 to 1500 in increments of 100 in column A, and the formulas (-0.01(A2+300)(A2-2300)) in B2 and (-0.01A2(A2-3000)) in C2. Remember, an asterisk (*) or a space must be used to show multiplication. Drag the formulas down to fill in the values for C($) and R($) for all your values of x, then write down:

(a) how much it would cost to produce the following number of calculators per day

(i) 100 (ii) 500 (iii) 1000 (iv) 1500

(b) how much revenue the company receives if it sells the following number of calculators per day

(i) 100 (ii) 500 (iii) 1000 (iv) 1500

The following screenshot shows the first few values you should obtain if you have entered your formulas correctly.

3 (a) Enter a profit title, Profit P($), in D1 and, using the fact that profit equals revenue minus cost, a formula in D2 to calculate the profit from the sale of x calculators.

(b) Show (i) by using the spreadsheet and (ii) by using algebra that this profit formula is P($) = 10x − 6900.

Technology


Technology Exploration GeoGebra


4 Produce a graph of P = 10x − 6900 for x-values between 0 and 1500. To do this, we need to first set up the axes and the range of values for both the x- and y-axes. From the menu, select Options and Drawing Pad. For the x-axis, input distance as 200, select x from the label list and input -200 as the min and 2100 as the max.

For the y-axis, input 1000 for the distance, y as the label, -9000 for the min and 16 000 for the max.

Input the formula in the Input bar at the bottom of your screen.

Use f(x)=Function[10x-6900,0,1500] to restrict the graph to the required values.

Use the text box tool to label your graph P($) = 10x − 6900.

5 What will the selling price of a calculator be if 1500 calculators are made each day?

6 What profit will be made on each calculator if 1500 calculators are made and sold each day?

7 Mr Slew recommends that the company introduces a new process to increase daily production to 2000 calculators by buying some new equipment. The cost of producing x calculators is now: D($) = -0.01(x + 400)(x − 2200), x ≤ 2000.

8 Enter the new cost title D($) into E1 and the new profit title $N into F1. Enter the new cost formula into to E2 and the new profit formula into F2 to calculate the profit N($) using Mr Slew’s new process.

9 Use algebra to simplify this rule.

10 Produce a graph of this rule using the same values of x as in Question 2. (You will need to input this rule with a new name, for example g(x).)

11 What does the point of intersection of the two graphs tell you?

12 As the manager of the company, would you introduce this new process? If so, how many calculators would you need to produce and sell each day to make the changeover to the new process worthwhile (i.e. to improve your profits)? Use your graphs or algebra to justify your decision.

13 Do the graphs help you understand what is happening to the profit in both the old and new processes? Explain why or why not.

Taking it further14 Use your spreadsheet and a cost equation in the

form of F(x) = -0.01(x + a)(x + b) to find a profit equation. Choose various values for a and b to find the maximum profit you could make on the manufacture and sale of 1500 calculators.


3 Algebra 177

Expanding special productsDid you notice anything interesting about your answers to Questions 5 and 6 in the last section (page 173)? Was there a pattern that could be used to expand using rules? All the parts of Question 5 were the square of sums or differences of two terms. All the parts of Question 6 were the product of the sum and difference of two terms.

Here we will investigate the expansion of (a + b)(c + d) for two special products.

Perfect squaresWhen we square a sum or a difference of two terms, we call this a perfect square. We expand perfect squares in exactly the same way as other binomial products.

Square of a sum Square of a difference

(a + b)2 = (a + b)(a + b) or (a − b)2 = (a − b)(a − b) = a2 + ab + ab + b2 = a2 − ab − ab + b2

= a2 + 2ab + b2 = a2 − 2ab + b2

This gives us a rule for expanding perfect squares

(a + b)2 = a2 + 2ab + b2

First Second First Twice the Secondterm term term product of term(a) (b) squared the two terms squared

(a2) (2 × a × b) (b2)

(a − b)2 = a2 − 2ab + b2

Perfect Square rules

(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

Note that the only difference in these two rules is the sign of the middle term and that this sign is the same as the sign inside the brackets.

Worked Example 19Expand each of the following expressions using a Perfect Square rule:

(a) (p + 4)2 (b) (d − 3)2 (c) (2f + 5)2 (d) (3m − 2n)2

Thinking(a) Use the rule (a + b)2 = a2 + 2ab + b2.

(Substitute a = p and b = 4.)(a) (p + 4)2 = p2 + 2 × p × 4 + 16

= p2 + 8p + 16

(b) Use the rule (a − b)2 = a2 − 2ab + b2. (Substitute a = d and b = 3.)

(b) (d − 3)2= d2 − 2 × d × 3 + 9= d2 − 6d + 9

(c) Use the rule (a + b)2 = a2 + 2ab + b2. (Substitute a = 2f and b = 5.)

(c) (2f + 5)2 = (2f)2 + 2 × 2f × 5 + 25= 4f 2 + 20f + 25

19


3.6


Difference of two squaresWhen we multiply the sum of two terms by the difference of those terms, we see that the terms inside the brackets are the same but the signs are opposite. This gives us another useful rule.

(a + b)(a − b) = a2 − ab + ab + b2

= a2 − b2

This rule is often referred to as DOTS (Difference of Two Squares).

Expanding special products

Fluency1 Expand each of the following expressions using a Perfect Square rule.

(a) (d − 1)2 (b) (m + 5)2 (c) (k + 2)2

(d) (p − 3)2 (e) (x − 8)2 (f) (y − 12)2

(g) (m + n)2 (h) (w − k)2 (i) (6 − c)2

(d) Use the rule (a − b)2 = a2 − 2ab + b2. (Substitute a = 3m and b = 2n.)

(d) (3m − 2n)2

= (3m)2 − 2 × 3m × 2n + 2n × 2n= 9m2 − 12mn + 4n2

Difference of Two Squares rule

(a + b)(a − b) = a2 − b2

Worked Example 20Use the Difference of Two Squares rule to expand the following:

(a) (x + 3)(x − 3) (b) (5 − k)(5 + k) (c) (3x + 2)(3x − 2) (d) (4m − 5n)(4m + 5n)

Thinking(a) Use the rule (a + b)(a − b) = a2 − b2.

(Substitute a = x and b = 3.)(a) (x + 3)(x − 3) = x2 − 32 = x2 − 9

(b) Use the rule (a + b)(a − b) = (a − b)(a + b) = a2 − b2. (Substitute a = 5 and b = k.)

(b) (5 − k)(5 + k) = 52 − k2 = 25 − k2

(c) Use the rule (a + b)(a − b) = a2 − b2. (Substitute a = 3x and b = 2.)

(c) (3x + 2)(3x − 2) = (3x)2 − (2)2 = 9x2 − 4

(d) Use the rule (a − b)(a + -b) = a2 − b2. (Substitute a = 4m and b = 5n.)

(d) (4m − 5n)(4m + 5n) = (4m)2 − (5n)2

= 16m2 − 25n2

NavigatorQ1 Column 1, Q2 Column 1, Q3, Q4, Q5, Q6, Q7, Q9 Column 1, Q11 (a), Q12

Q1 Column 2, Q2 Column 2, Q3, Q4, Q5, Q6, Q7, Q8, Q9 Columns 1&2, Q11, Q12

Q1 Column 3, Q2 Column 3, Q3, Q4, Q5, Q6, Q7, Q8, Q9, Q10, Q11, Q12

20

3.6

Answerspage 610

19


3

Algebra

3

.

6

179

(j)

(

y

+

6

)(

y

+

6

)

(k)

(

x

+

9

)(

x

+

9

)

(l)

(

k

−

7

)(

k

−

7

)

(m)

(

4

y

+

3

)

2

(n)

(

2

w

+

9

)

2

(o)

(

3

x

+

5

)

2

(p)

(

6

k

−

1

)

2

(q)

(

7

−

2

a

)

2

(r)

(

3

−

8

d

)

2

(s)

(

1

−

4

c

)(

1

−

4

c

)

(t)

(

6

+

7

y

)(

6

+

7

y

)

(u)

(

4

g

−

h

)(

4

g

−

h

)

(v)

(

3

a

+

d

)

2

(w)

(

2

a

−

3

b

)

2

(x)

(

5

m

+

8

n

)

2

2

Use the Difference of Two Squares rule to expand the following.

(a)

(

c

−

5

)(

c

+

5

)

(b)

(

a

+

4

)(

a

−

4

)

(c)

(

k

+

8

)(

k

−

8

)

(d)

(

9

+

x

)(

9

−

x

)

(e)

(

3

−

y

)(

3

+

y

)

(f)

(

1

−

k

)(

1

+

k

)

(g)

(

m

−

n

)(

m

+

n

)

(h)

(

x

−

y

)(

x

+

y

)

(i)

(

c

+

d

)(

c

−

d

)

(j)

(

v

+

w

)(

v

−

w

)

(k)

(

p

+ q)(p − q) (l) (h − g)(h + g)

(m) (2m − 1)(2m + 1) (n) (3x + 4)(3x − 4) (o) (6b − 7)(6b + 7)

(p) (1 + 5y)(1 − 5y) (q) (8 − 5a)(8 + 5a) (r) (3 − 2d)(3 + 2d)

(s) (4a + 5c)(4a − 5c) (t) (3m − 2n)(3m + 2n) (u) (2x − 7y)(2x + 7y)

(v) (4 + ab)(4 − ab) (w) (3 − cd)(3 + cd) (x) (5 + jk)(5 − jk)

3 (a + 5)2 is the same as:

A a2 + 5a + 10 B a2 + 10a + 25 C a2 + 10a + 10 D a2 + 25

4 (2k − 7)2 is the same as:

A 2k2 − 28k + 49 B 4k2 + 28k + 49 C 4k2 − 49 D 4k2 − 28k + 49

5 When expanded, (2 − 4x)(2 + 4x) is:

A 4 − 16x2 B 4 + 8x C 4 − 8x2 D 4 − 8x

6 When expanded, (ab + 3cd)(ab − 3cd) is:

A ab − 9cd B a2b2 + 9c2d2 C a2b2 − 9c2d2 D a2b2 + 6c2d2

Understanding7 (a) Find the area, in terms of x, of each of these squares. Expand your expressions.

(i) (ii)

(b) Calculate the area of each when x = 4 cm.

8 A square playing field is to be changed to a rectangular shape by adding 2 m to the length and subtracting 2 m from the width.

(a) Write an expression for the area of the square playing field.

(b) Write an expression for the area of the new rectangular playing field.

(c) Which playing field has the larger area? How much more area is available?

20

x + 7 2x – 3

x

x

x

x – 2

x

2

2


3.6


Reasoning9 Because 16 can be written as (10 + 6) or (20 − 4), 162 can be written as (10 + 6)2 or (20 − 4)2.

Evaluate the following, without a calculator, using your knowledge of special products.

(a) 232 (b) 492 (c) 1122

(d) 3.82 (e) 6.012 (f) 11.962

(g) 102 × 98 (h) 2.3 × 1.7 (i) 6.04 × 5.96

10 A steel washer has the dimensions shown.

(a) Find the area of one side of the washer by using your knowledge of the expansion of special products.

(b) What can you say about the area of the washer for any value of a where a > 0?

11 (a) Find the shaded area using:

Area of a trapezium = (a + b)h

(b) Find the shaded area using:

Area of a triangle = bh

(c) Write an equation using part (a) and part (b).

(d) If L = 5l, find the value of the shaded area:

(i) using your answer to part (a)

(ii) using your answer to part (b).

(e) Show that the shaded area is 96% of the area of the large triangle when L = 5l.

Open-ended12 Find three values for a and b so that a2 − b2 produces a square number.

Inner diameter

= a – 1a

Outer diameter

= a + 1a

l cm

L cm

L

l

cm

cm12---

12---

Problem solvingOranges and applesThe aim is to end up with the oranges and apples in opposite positions. You can make two types of moves:

• You can slide a piece of fruit one square into an empty space

• You can hop a piece of fruit over a different type of fruit

1 What is the minimum number of moves needed so that the oranges are on the left and the apples are on the right?

2 If there were four or five pieces of each fruit, how many moves would it take?

• Solve a simpler problem.

• Act it out.

• Look for a pattern.

Strategy options


3 Algebra 181

Factorising using common factorsThe opposite or inverse of expanding is called factorising.

Expanding Factorisinga(b + c) = ab + ac ab + ac = a(b + c)

There are many different ways to factorise algebraic expressions, but in this section we are going to find all the factors that are common to all terms in the expression. This is the highest common factor (HCF).

This procedure is called ‘taking out the highest common factor’. The HCF may be any positive number, a variable or the product of a number and one or more variables. Once an expression is factorised, no matter what method has been used, it will have brackets in it.

Factorising using the HCF

• Place the HCF outside brackets.

• Divide each term by the HCF; this quotient goes inside the brackets.

• If the HCF is not taken out as a factor, the expression is not fully factorised (e.g. 4x + 8 = 2(2x + 4) is not fully factorised).

• A negative sign may be taken out as well as the HCF; be careful to change signs inside the brackets (e.g. -5x + 10 = -5(x − 2)).

Worked Example 21Factorise each of the following:

(a) 5x + 20 (b) y2 − y (c) -9y − 6z + 3

(d) 3a2bc − acd (e) -8mn2 − 12m2n (f) a(b + 3) + 2(b + 3)

Thinking(a) Find the HCF of both terms (HCF = 5).

Write it down and follow with brackets. Divide each term by the HCF and write this quotient inside the brackets.

(a) 5x + 20 = 5(x + 4)

(b) Find the HCF of both terms (HCF = y). Write it down and follow with brackets. Divide each term by the HCF and write this quotient inside the brackets.

(b) y2 − y = y(y − 1)

(c) Find the HCF of both terms (HCF = 3). Write it down and follow with brackets. Take out the negative sign also. Divide each term by the factor outside the brackets and write the quotient inside the brackets. Remember to change the signs inside the brackets.

(c) -9y − 6z + 3 = -3(3y + 2z − 1)

21


3.7


Factorising using common factors

Fluency1 Factorise each of the following.

(a) 3x + 9 (b) 12y + 24 (c) 6 + 42a

(d) 2k − 6 (e) 4m − 20 (f) 8b − 4

(g) 7x − xy (h) 4s − 9st (i) 3a − ab

(j) 15y + 5w + 10x (k) 2a + 4b − 8c (l) 3x − 15y + 9

(m) 3ac + 2abc (n) 5xy2 + 2x2y (o) 9f 2gh − 12fg2

(p) 6gh2 + 18g2h (q) 16x2y − 40xy2 (r) 18ab − 42a2b

(s) -4m − 20 (t) -7d − 49 (u) -12r − 60

(v) x(a + 3) + 5(a + 3) (w) m(n − 2) + 9(n − 2) (x) 3(x + 5) + y(x + 5)

2 Factorise each of the following.

(a) 10x + 25 (b) 16y + 36 (c) 14a − 35

(d) 12d − 18 (e) 18 − 27m (f) 36 − 15k

(g) 24abc − 10b + 7bc (h) 5km − 3kn + 2ghk (i) c + 4bc + 2ac

(j) x2 + 3x (k) y2 + 6y (l) 5k + k2

(m) 8m − 3m2 (n) 4p2 − 7p (o) 9a2 − 5a

(p) 6d − 9d2 (q) 22g − 14g2 (r) 24a − 32a2

(d) Find the HCF of both terms (HCF = ac). Write it down and follow with brackets. Divide each term by the HCF and write this quotient inside the brackets.

(d) 3a2bc − acd = ac(3ab − d)

(e) Find the HCF of both terms (HCF = -4mn) Write it down and follow with brackets. Divide each term by the HCF and write this quotient inside the brackets. Remember to change the signs inside the brackets.

(e) -8mn2 − 12m2n = -4mn(2n + 3m)

(f) Find the HCF of both terms (HCF = (b + 3)). Write it down and follow with brackets. Divide each term by the HCF and write this quotient inside the brackets.

(f) a(b + 3) + 2(b + 3) = (b + 3)(a + 2)

NavigatorQ1 Column 1, Q2 Column 1, Q3 Column 1, Q4 Column 1, Q5, Q6, Q7, Q9, Q12, Q13

Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4 Column 1, Q5, Q6, Q8, Q9, Q10, Q12, Q13

Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4 Column 2, Q5, Q6, Q7, Q8, Q9, Q10, Q11, Q13

3.7

Answerspage 610

21


3

Algebra

3

.

7

183

3

Factorise each of the following by taking out the highest negative common factor.

(a)

-2

m

−

12

(b)

-4

k

−

24

(c)

-5

h

−

25

(d)

-

x

2

−

x

(e)

-

y

2

−

3

y

(f)

-

p

2

−

4

p

(g)

-16

a

+

4

(h)

-21

b

+

7

(i)

-30

r

+

6

(j)

-24

p

+

40

(k)

-18

w

+

16

(l)

-45

k

−

63

(m)

-15

x

2

−

36

x

(n)

-14

y

2

−

49

y

(o)

-20

r

2

−

44

r

(p)

-8

d

−

4

f

−

6

g

(q)

-15

a

−

10

b

−

5

c

(r)

-14

c

−

28

d

−

7

f

(s)

-12

x

2

−

16

x

+

20

xy

+

4

xz

(t)

-2

km

+

4

km

2

−

6

k

−

8

k

2

(u)

-10

b

2

+

4

bc

−

8

bc

2

+

16

ab

4

Factorise each of the following.

(a)

4

(

y

−

1

) +

w

(

y

−

1

)

(b)

5

(

m

−

3

) +

n

(

m

−

3

)

(c)

5

(

p

−

2

) +

r

(

p

−

2

)

(d)

p

(

q

+

5

) −

2

(

q

+

5

)

(e)

y

(

2

x

+

3

) − (

2

x

+

3

)

(f)

a

(

3

p

+

5

) − (

3

p

+

5

)

(g)

d

(

f

−

1

) −

6

(

f

−

1

)

(h)

5

d

(d − 2) − 4(d − 2) (i) 3m(m − 6) − 7(m − 6)

5 When factorised, the expression 3ab2c − 6abc + 3bc is:

A 3bc(ab − 2a) B 3abc(b − 2 + 3a)

C 3bc(ab − 2a + 1) D 6bc(ab − a + 2bc)

6 When fully factorised, the expression -2x(3m − 4) + 5y(3m − 4) is:

A (3m − 4)(5y − 2x) B (5y − 4)(3m − 2x)

C (-3m + 4)(-2x + 5y) D (-2x − 4)(3m + 5y)

Understanding7 (a) Write the expression for ‘Eight times the product of two

numbers added to twelve times one of the numbers’.

(b) Factorise the expression in (a).

8 A rectangular pizza is of length d cm and width(c + 2) cm. Another rectangular pizza of thesame width has a length of 5 cm. The two pizzas are joined to make a longer pizza.

(a) Write down the area of the new shape in factorised form.

(b) What is the relationship between c and d if the pizza is square?

9 A floor-to-ceiling window of width x m is to be installed in a wall 5 m long and (y + 3) m high. The remaining part of the wall is to be plastered.

(a) Find the length of the wall to be plastered after the window is installed.

(b) Find the area of the wall to be plastered after the window is installed.

Reasoning10 A wood template is to be made for the front of a letter box with the

dimensions shown.

(a) Write an expression that would show the area of timber used in the construction before the circular opening is cut out.

(b) Express this answer in a factorised form.

(c) If the diameter of the circular cut out needs to be 2 cm less than the width, find the area of the timber cut out for the opening.

(d) Write this in expanded form.2x cm

2x cm

y cm


3.7


(e) Show that the total area of timber that will be used is (4 − π)x2 + (y + 2π)x − π cm2.

(f) What area of timber would be wasted if the template was cut from a rectangular piece of timber with the smallest possible dimensions?

11 A lasagne is baked in a tray (3p + 2) cm long and (2p − 1) cm wide. A slice, the full width of the tray and (r + 2) long is taken from the lasagne.

(a) Find the area of the lasagne, before the slice was removed, in factorised form.

(b) Find the area of the slice taken from the lasagne in factorised form.

(c) Use your answers to parts (a) and (b) to find the area remaining after the slice has been removed. Write your answer in simplest factorised form.

(d) How could you find the area in part (c) more simply? Show your working.

Open-ended12 Make up three different expressions whose highest common factor is 3x.

13 In a test, students were asked ‘Factorise the expression -5x + 10’. Three students wrote these answers; one answer was correct, the other two answers were incorrect. Identify the correct answer and explain the mistakes the other two students made.

A -5(x + 2) B -5(x + 10) C -5(x − 2)

GameMatching expressionsEquipment required: 2 brains, 20 pieces of paper cut into rectangles the size of playing cards

How to win: Match the most expressions and get more cards than your opponent

How to play:

1 Each player writes out five expressions in expanded form, one on each card, and their equivalent factored form on another five cards.

2 All 20 cards are mixed up and laid out face down. This is now a game of memory.

3 Player 1 turns two cards over. Players have 10 seconds to work out whether the cards are a match.

4 If they are a match, Player 1 takes the cards and has another go, otherwise it is Player 2’s turn.

5 After all the cards have been matched and claimed by a player, the game ends and the player with the most cards is the winner.

AA Pearson Maths 9 SB-03.fm 4 Monday, October 24, 2011 10:19 AM

3 Algebra 185

Mr Gershwin’s pianoMr Gershwin wants to build a small performance theatre with a stage for piano recitals. The theatre cannot be longer than 16 metres, but must be able to seat at least 108 people in the main seating area, with extra space allowed outside the seating area to accommodate wheelchairs.

The theatre in which the stage is to be built is rectangular with the following specifications:

• Its width, W, in metres correct to one decimal place is given by the expression W = 4x − 4.

• Its length, L, in metres correct to one decimal place, is twice the theatre’s width.

The Big QuestionMr Gershwin wants to know the dimensions of the smallest and largest theatre he can build with the conditions given above. Can you help him?

Engage1 (a) Find L and W for x = 2.

(b) Draw a diagram of this theatre showing the dimensions given in part (a).

The stage is also rectangular and is to be built across the width of the theatre so that the length of the stage, in metres, a, is equal to the width of the theatre. The width of the stage, in metres, b, is half its length.

(c) Find the value of a and b for x = 2.

(d) Draw the stage in your diagram, showing the dimensions given in part (c).

ExploreThe audience is to be seated in rows in a rectangular space so that there is a 2-metre clearance from the front of the stage and from the back of the room and with a 1-metre aisle either side.

Investigation



2 (a) Draw the seating area in your diagram showing the clearances.

(b) Find the length in metres, c, and width in metres, d, of the audience space for x = 2.

(c) Add the dimensions of your seating area to your diagram.

(d) If a space of 50 cm × 50 cm is allowed for each seat, how many seats could fit into this space?

(e) Choose three other values for x and calculate the number of people that could be seated for each of these values. Do not use integer values only.

Explain3 (a) Explain why values for x ≤ 1 cannot be used for

the length and width of the theatre.

(b) What did you find happened to the number of people that could be seated when you changed the values of x?

(i) Explain why the smallest value for x is 1 m if there is to be any seating at all.

(ii) How many seats can fit if x = 1 m?

Elaborate4 (a) By substituting different values for x, find the

dimensions of the smallest theatre that Mr Gershwin can build so that 108 people can be seated in the seating area. Use your results from 2 (e) to help you.

(b) What is the largest value that x can be so that the length of the theatre does not exceed 16 metres?

(c) Hence, determine the range of values for x that Mr Gershwin could use to build his theatre.

Evaluate5 (a) Is it sensible to change the size of the stage as the

size of the theatre changes? Give a reason for your answer.

(b) Do you think the relationship used here is a good one? Give a reason for your answer.

ExtendMr Gershwin also wants to know how much carpet he would need to cover the floor space not occupied by the stage or the seating area. You need to do the following calculations to find the area of floor to be covered with carpet.

6 (a) Find W and L in terms of x and factorise your answer.

(b) Find the area of the theatre in terms of x. Write your answer first in factorised form and then expand it.

(c) Find a and b in terms of x in factorised form.

(d) Find the area of the stage in terms of x. Write your answer first in factorised form and then expand it.

(e) Find c and d in terms of x.

(f) Find the seating area in terms of x. Write your answer first in factorised form and then expand it.

(g) Find the area in terms of x of the floor space of the theatre occupied by both the stage and the seating area.

(h) Use the area equations found in parts 6 (b), (d) and (f) to show that the area, in terms of x, of the remaining floor space in the theatre is equal to (28x − 36) m2.

(i) Show how this area could be found using an easier method.

• Make a table.

• Draw a diagram.

• Guess and check.

• Break the problem into manageable parts.

Strategy options

38---

38---


3 Algebra 187

Factorising by grouping in pairsIf there are four terms in an expression and there is no common factor for all four terms, we can sometimes group the expression in pairs of terms so that there is a common factor in each pair. If this gives us another common factor in brackets, we can factorise further. This technique is called grouping in pairs. Sometimes you may have to consider which two terms to pair together.

Worked Example 22Factorise the following:

(a) ab + 5b + 3a + 15 (b) 3x + 12 − xy − 4y

Thinking(a) 1 Factorise by taking a common factor

from the first pair of terms (b) and a different common factor from the second pair of terms (3).

(a) ab + 5b + 3a + 15= b(a + 5) + 3(a + 5)

2 Now we have a new common factor. (Here (a + 5).) Write this factor in front of another set of brackets. Divide each product by this factor and write the quotient in the second set of brackets.

= (a + 5)(b + 3)

(b) 1 Factorise by taking a common factor from the first pair of terms (3) and a different common factor from the second pair of terms (-y). When taking out a negative common factor, change the signs inside the brackets.

(b) 3x + 12 − xy − 4y= 3(x + 4) − y(x + 4)

2 Now we have a new common factor. (Here (x + 4).) Take out this common factor to complete the factorisation.

= (x + 4)(3 − y)

Worked Example 23Factorise df + 6g − 6f − dg

Thinking1 The first two terms (df and 6g) do not

have a common factor. So, rewrite the expression to pair terms such that each pair of terms does have a common factor.

df + 6g − 6f − dg = df − 6f − dg + 6g

22

23


3.8


In Questions 4, 5 and 6 of Exercise 3.5 (page 173), we expanded some products that gave us quadratic expressions when we simplified. We explored expanding some special products in Exercise 3.6 and these also gave us quadratic expressions. Because factorising and expanding are opposite operations, we can use them to factorise quadratics. We will use our knowledge of factorising by grouping in pairs to show how this process works.

2 Now factorise the first pair of terms by taking out a common factor (f) and take a different common factor from the second pair (-g). Change signs if necessary.

= f(d − 6) − g(d − 6)

3 Take out the remaining common factor to complete the factorisation. (Here (d − 6).)

= (d − 6)(f − g)

Worked Example 24For each of the following expressions, (i) factorise, (ii) simplify, (iii) write the relationship between your answers to parts (i) and (ii), and (iv) state the relationship between the coefficients of the two middle terms and the last term.

(a) x2 + 5x + 6x + 30 (b) p2 − 4p − 4p + 16 (c) a2 − 12a + 12a − 144

Thinking(a) (i) 1 Factorise by taking out a common

factor from the first two terms and by taking out a different factor from the next two terms.

(a) (i) x2 + 5x + 6x + 30= x(x + 5) + 6(x + 5)

2 Factorise by taking out the new common factor.

= (x + 5)(x + 6)

(ii) Add the two middle terms because they are like terms.

(ii) x2 + 5x + 6x + 30= x2 + 11x + 30

(iii) These answers must be equal to each other because they are both equal to the expression.

(iii) x2 + 11x + 30 = (x + 5)(x + 6)

(iv) 1 Identify the coefficients of the two middle terms, remembering to include the signs, and look for a relationship with the last term (the product of these coefficients gives the last term).

(iv) 5 × 6 = 30

2 State the relationship you have found.

The product of the coefficients of the two middle terms gives the last term.

24


3 Algebra

3.8

189

(b) (i) 1 Factorise by taking out a common factor from the first two terms and by taking out a different factor from the next two terms.

(b) (i) p2 − 4p − 4p + 16= p(p − 4) − 4(p − 4)


= (p − 4)(p − 4)

3 Write as a perfect square. = (p − 4)2


(ii) p2 − 4p − 4p + 16= p2 − 8p + 16


(iii) (p − 4)2 = p2 − 8p + 16

(iv) 1 Identify the coefficients of the two middle terms, remembering to include the signs, and look for a relationship with the last term (the product of these coefficients gives the last term).

(iv) -4 × -4 = 16



(c) (i) 1 Factorise by taking out a common factor from the first two terms and by taking out a different factor from the next two terms.

(c) (i) a2 − 12a + 12a − 144= a(a − 12) + 12(a − 12)


= (a − 12)(a + 12)


(ii) a2 − 12a + 12a − 144= a2 − 144


(iii) a2 − 144 = (a − 12)(a + 12)

(c) (i) 1 Identify the coefficients of the two middle terms, remembering to include the signs, and look for a relationship with the last term (the product of these coefficients gives the last term).

(c) (i) -12 × +12 = -144




3.8


Factorising by grouping in pairs

Fluency1 Factorise the following.

(a) xy + 4y + 3x + 12 (b) pq + q + 5p + 5 (c) mn + 2n + 9m + 18

(d) ab + bf + ad + df (e) pq + qt + pr + rt (f) km + 3nk + 6m + 18n

(g) np − 7n + p − 7 (h) ab − 5b + a − 5 (i) mn − 2n + 7m − 14

(j) ab − 6b − 4a + 24 (k) x2 + 2x − 8x − 16 (l) y2 + 4y − 6y − 24

2 Factorise the following.

(a) 2p + 3k + kp + 6 (b) mn + pq + np + mq (c) cd + 12 + 6c + 2d

(d) ad − hk − dk + ah (e) xy − 8 − 4y + 2x (f) eg − 3 − g + 3e

(g) bc − 1 + c − b (h) xy − 3 + 3x − y (i) mk − 3n + 3m − nk

3 For each of the following expressons (i) factorise, (ii) simplify, (iii) write the relationship between your answers to parts (i) and (ii), and (iv) state the relationship between the coefficients of the two middle terms and the last term.

(a) x2 + 7x + 2x + 14 (b) j2 + 3j + 5j + 15 (c) z2 + z + 8z + 8

(d) x2 + 2x − 8x − 16 (e) y2 + 4y − 6y − 24 (f) x2 + 3x − 4x − 12

(g) a2 − 4a + 5a − 20 (h) k2 − 2k + 5k − 10 (i) m2 − 7m + 4m − 28

(j) y2 − 3y − 7y + 21 (k) c2 − 9c − 5c + 45 (l) d2 − 8d − 12d + 96

(m) m2 + 3m + 3m + 9 (n) r2 + 6r + 6r + 36 (o) k2 + 11k + 11k + 121

(p) p2 − 5p − 5p + 25 (q) d2 − 7d − 7d + 49 (r) n2 − 8n − 8n + 64

(s) c2 − 3c + 3c − 9 (t) t2 − 2t + 2t − 4 (u) b2 − 9b + 9b − 81

4 By taking out the common factor from k( j − 1) + ( j − 1) we obtain:

A ( j − 1)(k + 1) B ( j + 1)(k − 1) C ( j − 1)(k) D ( j + 1)(k + 1)

Understanding5 Square a number, x, subtract four times x, add 12 and subtract three times x.

(a) Write down the expression for the information given.

(b) Simply the expression in (a).

(c) Factorise the expression in (a).

6

(a) Write down the expression for the total area of the four shapes above.

(b) Factorise the expression in (a).

NavigatorQ1 Column 1, Q2 Column 1, Q3 Column 1, Q4, Q5, Q6, Q7, Q11

Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4, Q5, Q6, Q7, Q11, Q12

Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4, Q5, Q6, Q7, Q8, Q9, Q10, Q11, Q12

3.8

Answerspage 611

22

23

24

y

y

y

6

y

6 6

6


3

Algebra

3

.

8

191

(c)

Draw a diagram to show how the four shapes can fit together to make a square (include dimensions on your diagram).

Reasoning

7 (a)

Factorise

ab

+

ac

+

ad

+

7

b

+

7

c

+

7

d

by grouping in threes. That is, take out a common factor from three terms at a time.

(b)

Show that, if

b

=

a

and

c

+

d

=

7, the expression is a perfect square.

8 (a)

Rewrite 7

x

using two terms so that the expression

x

2

+

7

x

+

10 can be factorised by grouping in pairs.

(b)

Now factorise your new expression.

9 (a)

Factorise 35

y

2

−

21

y

.

(b)

If this represents the area of a rectangle in cm

2

and the factors represent the width, write down the dimensions of the rectangle.

(c)

If an area of 15

y

−

9 cm

2

is added to the area in part

(a)

, find the combined area in factorised form.

(d)

Write down the dimensions of the new rectangle if the factors in part

(c)

represent the length and width.

10

A trapezium has an area of m

2

where

x

and

y

are both positive numbers and

x

>

2.

(a)

Write the area in factorised form.

(b)

If the smaller factor represents the height of the trapezium, what is the sum of the lengths of the two parallel non-equal sides?

Open-ended

11 (a)

Find two terms that, when added to

x

2

+

ax

, can be factorised by grouping in pairs.

(b)

Factorise your expression in

(a)

.

12 (a)

Draw a rectangle of length

l

and width

w

. Extend the length by a certain amount and the width by a different amount. Use your diagram to write the area of the rectangle in two different ways.

(b)

Draw a square of side

s

. Extend one pair of opposite sides by a certain amount and the pair of opposite sides by a different amount. Use your diagram to write the area of the new rectangle in two different ways.

(c)

Draw a square of side

s

and extend the sides by the same amount to form a new square. Use your diagram to find the area of the new square in two different ways.

3x2 2xy 6x 4y––+2

------------------------------------------------

PuzzleTrouble in variable paradiseSome variables were messaging online; the following is an excerpt of their conversation…

x: I am like soooo smaller than u y

y: Lol, yeah but at least I’m not negative

x: *cries* u might be greater than me, but at least I’m not square like u

y: Aw, don’t be cut x, together we make one, anyway although you’re negative, you’re a perfect cube.

x: Luv u y

y: cul8r x

What are the values of x and y?


Challenge 3

192

PEARSON

mathematics 9

1

Write two numbers that have a sum of 27 and a product of 72.

2

Bridget once said, ‘The day before yesterday I was 15 and next year I will be 18’. When is Bridget’s birthday and on what day did she say this?

3

How many perfect cubes are there between 2000 and 10 000?

4

Part of a rectangle is shaded as shown in the diagram.

Which expression gives the area of the shaded part?

A

2

x

2

−

x

−

8

B

2

x

2

−

x

−

12

C

2

x

2

−

3

x

−

8

D

2

x

2

−

3

x

−

12

5

If

♦

represents a digit, find the value of

♦

if [3

(

170

+

♦)

]

2

=

272 48

♦

.

6

a

,

b

and

c

are positive whole numbers. Show that

(

a

–

b

)(

b

–

c

)(

c

–

a

)

is divisible by 2.

7

Is the number 3

404

+

4

495

divisible by 5? Show why or why not.

8

Steven thinks of four different numbers. The sums Steven gets by pairing the numbers are 6, 7, 9, 9, 11 and 12. What are the four numbers?

9

A formula used in the science of optics is

=

(a)

Make

f

the subject of the formula.

(b)

Make

v

the subject of the formula.

10 (a)

Show that the difference between any two-digit number and the number formed by reversing the digits is divisible by 9. (

Hint

: let the two-digit number be 10

a

+

b

).

(b) The difference between any three-digit number and the number formed by reversing its digits is obtained. By what number is it divisible?

(c) The difference between any four-digit number and the number formed by reversing its digits is obtained. By what number is it divisible?

6

(x – 2)

2

5

(2x + 1)

(Diagram not to scale)

1u--- 1

v---+ 1

f--- .


1933 Algebra

Chapter review D.I.Y. Summary

Copy and complete the following using the words and phrases from the list where appropriate to write a summary for this chapter. A word or phrase may be used more than once.

1 An expression of the form x2 − y2 is called the .

2 If an expression is factorised to obtain the product of two brackets that are exactly the same, it is called a .

3 To an expression is to remove brackets. The reverse operation is to the expression and put it into brackets.

4 If we write a number in , then it has a number between 1 and 10 multiplied by a positive or negative power of 10.

Fluency1 Simplify each of the following.

(a) a3b5 × a2b2 (b) m4n7 × m3n (c)

(d) (e) (f)

2 (2x3y2)3 is the same as:

A 2x3y6 B 8x3y6 C 2x9y6 D 8x9y6

3 Express each of the following with positive integers.

(a) -5m−1 (b) a2b−6 (c) x−2y−3 (d) a−4b−9

4 Write each of the following in scientific notation.

(a) 0.000 000 000 000 002 mm (b) 0.000 15 kg

(c) 320 000 000 L (d) 480 000 000 000 000 km

5 Write each of the following as a basic numeral.

(a) 5.4 × 107 (b) 8.273 × 10-4

(c) 1.4 × 10-5 (d) 7.162 × 103

6 Rearrange each of the following to make x the subject.

(a) z = (b) x + 2a = 4x

(c) s = xt + 8t2 (d) = 10

base expand highest common factor reciprocal

binomial expanded form index scientific notation

Difference of Two Squares expression index form significant figures

distributive law factorising perfect square variable

equation grouping in pairs quadratic trinomial

3Key Words

3.1p4q6

p3q9-----------

x5( )2 x8×x4( )4

-----------------------x2( )3 y4( )2×

xy3 x5×------------------------------- c2d4

c4d2---------- c6d

cd( )2------------×

3.2

3.2

3.3

3.3

3.43 xy–

2--------------

2 4z+3x

--------------



7 Expanded and simplified, 2c(3b − 4) − 3b(-5 + 2c) is:

A 15b − 8c B 21b + 8c C 12bc − 12c D 7b + 14c

8 Expand:

(a) 4(x + 7) (b) x(x − 3) (c) -5x(2x − 1)

9 (a) Expand and simplify:

(i) 4(2x − 3) + 5(x + 1) (ii) 3x(6x + 5) − 2(3 − 2x)

(b) Find the value of each expression when x has the value 4.

10 Expand and simplify:

(a) (x + 8)(x + 2) (b) (2a + 5)(3a − 4)

11 Expand and simplify:

(a) (x + 12)2 (b) (5c − 2d)2 (c) (4x − 3)(4x + 3)

12 Factorise:

(a) 8a + 12 (b) -15k − 20k2

(c) 5p2q − 3pq + 2pq2 − pqr (d) -24a − 6ab − 72

13 Factorise by grouping in pairs.

(a) ab + 4b + 3a + 12 (b) ab − 2b + a − 2

(c) ab − 6a − 5b + 30 (d) ab − 7 + 7b − a

Understanding14 Express the following in scientific form.

(a) A 10-cent coin is about 0.0015 m thick.

(b) Pluto is approximately 5900 million km from the Sun.

15 Express each of the following as a basic numeral.

(a) The distance from the Sun to Jupiter is approximately 7.78 × 108 km.

(b) The temperature at the centre of the Sun is believed to be about 1.5 × 107°C.

16 The length of a greeting card is 20 cm and the width is (5 + x) cm.

(a) Draw a diagram of the card.

(b) Write an expression for the perimeter.

(c) Factorise the expression obtained in part (b).

(d) Write an expression for the area (in factorised form).

(e) Expand the expression obtained in part (d).

(f) Check that the two expressions obtained for the area are equivalent by substituting a value for x.

17 If the area of a rectangular postage stamp is 4x2 + 12x, what is the width of the stamp if the height is 2x?

18 The mass of an object can be determined by finding the product of its density and its volume. If the density can be represented by (m − 2) and the volume by (k + 8), write an expression for the mass.

Suggest possible values that m and k could take to give a mass of 48.

19 Find the height of this triangular support structure if the base is 14y and the area enclosed by the beams is 35y2 − 21y.

3.5

3.5

3.5

3.5

3.6

3.7

3.8

14y


1953 Algebra

Reasoning20 The heart pumps 7000 L of blood daily.

(a) Express this quantity in scientific notation.

(b) How many litres of blood are pumped by the heart:

(i) each week?

(ii) each second, to one decimal place?

(iii) in 3 years (assuming no leap years)?

(c) Would this measurement be exact and the same for each person?

21 The Greenwood family had a rectangular block of land 2 km by 5 km for grazing sheep, as seen in the diagram. They then bought the land adjoining this leading down to the river. Depending on the time of year and the weather conditions, the width of the river can vary, so the land available for grazing sheep changes.

(a) Write an expression for the total area available for grazing sheep at any particular time. (Assume the land remains rectangular.)

(b) Suppose there are x sheep per km2 of land. How many sheep are there altogether?

Write your answer in expanded form.

(c) Fourteen sheep are moved to another location.

(i) Write the expression for the number of sheep remaining.

(ii) Expand (x + 7)(x − 2). How is this related to the expression in (i)?

(iii) Explain what happens when x = 2 km.

NAPLAN practice 3Numeracy: Calculator allowed1 Which of the following numbers is the largest?

A 3.2 × 1018 B 9.6 × 103 C 3.2 × 10-18 D 9.6 × 10-3

2 What is the value of 7(x − 9), when x = 15?

3 Factorise the expression 14x + 21.

Numeracy: Non-calculator4 Which one of the following expressions is equivalent to 3(5x − 2)?

A 15x − 1 B 15x − 6 C 15x + 1 D 15x + 6

5 A rule for y in terms of x is y = 12 − 3x. Write an equivalent rule for x in terms of y.

6 Which one of the following expressions is a fully factorised form of 9xy + 12y?

A 3(3xy + 4y) B 9y(x + 2) C y(9x + 12) D 3y(3x + 4)

7 When the expression -3x2y + 6xy is fully factorised, the result is:

A 3xy(2 − x) B -3xy(2 − x) C -3xy(x − y) D -3xy(x + y)

2 km

5 km

river

x

12---


aa pearson maths 9 sb-03.fm page 167 monday, october 24...

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